Definition 1.

For any two distinct elements x,y of P, there
exists exactly one element of L which contains both x and y.

Theorem 1.

If (P,L) is an abstract Hesse configuration,
then L has 12 elements.

Proof.

Let D be the set of all two-element subsets of P.
Then D has (92)=36 elements. Each
element of L is a subset of P with three elements,
hence has (32)=3 subsets of cardinality 2.
By the definition above, every element of D must
be a subset of exactly one element of L. For this
to be possible, L must have cardinality 36/3=12.
∎

Theorem 2.

If (P,L) is an abstract Hesse configuration then, for
every p∈P, there exist exactly four elements
ℓ∈L such that p∈ℓ.

Proof.

To every q∈P such that q≠p, there exists
exactly one ℓ∈L such that p∈ℓ and
q∈ℓ. Furthermore, for every ℓ∈L such
that p∈ℓ, there will be exactly two elements
of ℓ other than p. Hence, there exist (9-1)/2=4
elements ℓ∈L such that p∈ℓ.
∎

Theorem 3.

If (P,L) is an abstract Hesse configuration and ℓ∈L,
then there exist m,n∈L such that ℓ∩m=ℓ∩n=m∩n=∅.

Proof.

By the foregoing result, given p∈ℓ, there are four
elements of L to which p belongs. One of these, of course,
is ℓ itself, and the other three are distinct from ℓ.
Since ℓ has three elements, this means that there are at
most 3⋅3+1=10 elements k∈L such that
P∩L≠∅. Because L has 12 elements. there
must exist m,n∈L such that ℓ∩m=ℓ∩n=∅.

It remains to show that m∩n=∅. Suppose to the
contrary that there exists a p such that p∈m and
p∈n. Since m∩ℓ=∅, it follows that
p∉ℓ, hence there will exist three distinct elements
of L containing p and an element of ℓ. Because
ℓ∩m=ℓ∩n=∅, these three elements
are distinct from m and n. That makes for a total of
five distinct elements of L containing p, which
contradicts the previous theorem, hence m∩n=∅.
∎

Theorem 4.

If m,n,k are elements of L such that m∩n=n∩k=k∩m=∅ and ℓ∈L∖{m,n,k},
then ℓ has exactly one element in common with each of m,n,k.

Proof.

Since each element of L is a subset of P with three elements
and m,n,k are pairwise disjoint but P only has nine
elements, it follows that every element of P must belong
to exactly one of m,n,k. In particular, this means that
every element of ℓ must belong to one of m,n,k. Were
two elements of ℓ to belong to the same element of
{m,n,k} then, by the third defining property, that element
would have to equal ℓ, contrary to its definition. Hence,
each element of ℓ must belong to a distinct element of
{m,n,k}.
∎

Theorem 5.

If (P,L) is an abstract Hesse configuration, then we can
label the elements of P as A,B,C,D,E,F,G,H,I in such a way
that the elements of L are

{A,B,C},{D,E,F},{G,H,I},

{A,D,G},{B,E,H},{C,F,I},

{D,H,C},{A,E,I},{B,F,G},

{B,D,I},{C,E,G},{A,F,H}.

Proof.

By theorem 3, there exist a,b,c∈L such that
a∩b=b∩c=c∩a=∅. Since L has
twelve elements, there must exist an a elemetn of L distinct
from a,b,c. Pick such an element and call it d. By
another application of theorem 3, there must exist e,f∈L
such that d∩e=e∩f=f∩d=∅.

By theorem 4, a must have exactly one element in common with
each of d,e,f; let A the element it has in common with d,
B be the element it has in common with e and C be the
element it has in common with f. Likewise, b must have
exactly one element in common with each of d,e,f, as must c.
Let D be the element b has in common with d, E be the
element b has in common with e, F be the element b
has in common with f, G be the element c has in common
with d, H be the element c has in common with e and I
be the element c has in common with f.

Summarizing what we just said another way, we have assigned
labels A,B,C,D,E,F,G,H,I to the elements of P in such
a way that

a={A,B,C},b={D,E,F},c={G,H,I},

d={A,D,G},e={B,E,H},f={C,F,I}.

That is half of what we set out to do; we must still label
the remaining six elements of L.

By theorem 4, if ℓ∈L∖{a,b,c,d,e,f}, then
ℓ must have exactly one element in common with each of
a,b,c and exactly one element in common with each of d,e,f.

Suppose that A∈ℓ. It could not be the case that
D∈ℓ because then ℓ would have two elements in
common with d. Since ℓ must have one element in common
with b, that means that either E∈ℓ or F∈ℓ.
If A,E∈ℓ, then the element ℓ has in common with
c cannot be G because ℓ would have both A and G
in common with c and it cannot be H because ℓ and
e would have both E and H in common, hence the only
possibility is to have I∈ℓ, i.e. ℓ={A,E,I}.
Likewise, if A,F∈ℓ, it follows that H∈ℓ.

Summarrizing the last few sentences, if A∈ℓ, then
either ℓ={A,E,I} or ℓ={A,F,H}. By a similar
line of reasoning, if B∈ℓ, then either ℓ={B,D,I} or ℓ={B,F,G} and, if B∈ℓ, then
either ℓ={C,D,H} or ℓ={C,E,G}. Since ℓ
must contain one of A,B,C, it follows that there are omnly
the following six possibilities for ℓ:

{D,H,C},{A,E,I},{B,F,G},

{B,D,I},{C,E,G},{A,F,H}.

However, since L∖{a,b,c,d,e,f} has cardinality
six, all these possibilities must be actual members of the set.
∎